Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<iomanip>
#include<limits>
#include<thread>
#include<utility>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<queue>
#include<cmath>
#include<numeric>
#include<cassert>
#include<random>
#include<chrono>
#include<unordered_map>
#include<fstream>
#include<list>
#include<functional>
#include<bitset>
#include<complex>
using namespace std;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef pair<int,int> pi;
typedef pair<double,double> pd;
typedef pair<double,ll> pdl;
#define F first
#define S second
const ll E=1e18+7;
const ll MOD=1000000007;

ll n,m,s,t;
vector<vector<ll>> edge;

void dijk(ll s,vector<ll> &dp){
    dp.resize(n);
    for(auto &i:dp){i=E;}
    dp[s]=0;
    priority_queue<pll,vector<pll>,greater<pll>> Q;
    Q.push({0,s});
    while(!Q.empty()){
        pll w=Q.top(); Q.pop();
        if(dp[w.S]!=w.F){continue;}
        for(auto &i:edge[w.S]){
            if(dp[i]>dp[w.S]+1){
                dp[i]=dp[w.S]+1;
                Q.push({dp[i],i});
            }
        }
    }
}


int main(){
    cin>>n>>m>>s>>t;
    edge.resize(n);
    for(int i=0;i<m;i++){
        ll u,v;
        cin>>u>>v;
        u--; v--;
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    s--; t--;
    vector<ll> dp,dp2;
    dijk(s,dp);
    dijk(t,dp2);
    map<ll,ll> M,M2;
    for(auto &i:dp){M[i]++;}
    for(auto &i:dp2){M2[i]++;}
    ll ans=0;
    ll dist=dp[t];
    for(int i=0;i<n;i++){
        ans+=M[i]*M2[dist-i-2];
    }
    cout<<ans<<endl;
    
    
    
    return 0;
}